Q136 / 180 · KCET 2013 · Physics
For a transformed, the turns ratio is 3 and its efficiency is $0.75$. The current flowing in the primary coil is $2 \mathrm{~A}$ and the voltage applied to it is $100 \mathrm{~V}$. Then the voltage and the current flowing in the secondary coil are ... respectively
- A$150 \mathrm{~V}, 1.5 \mathrm{~A}$
- Correct answerB$300 \mathrm{~V}, 0.5 \mathrm{~A}$
- C$300 \mathrm{~V}, 1.5 \mathrm{~A}$
- D$150 \mathrm{~V}, 0.5 \mathrm{~A}$
Solution
Given, $\frac{n_{s}}{n_{P}}=3 ; \quad \eta=0.75 ; I_{P}=2 \mathrm{~A}$ and $V_{P}=100 \mathrm{~V}$
We know that,
$\begin{aligned}
&\qquad \eta=\frac{V_{S}}{V_{P}} \frac{I_{S}}{I_{P}} \\
&\text { or } \quad \eta=\left(\frac{n_{s}}{n_{P}}\right) \frac{I_{s}}{I_{P}} \quad\left(\because \frac{n_{s}}{n_{p}}=\frac{V_{s}}{V_{p}}\right) \\
&\Rightarrow \quad 0.75=\frac{3 \times I_{s}}{2} \\
&\Rightarrow \quad I_{s}=0.5 \mathrm{~A} \\
&\text { By the relation, } \frac{n_{s}}{n_{P}}=\frac{V_{s}}{V_{P}} \\
&\Rightarrow \quad 3=\frac{V_{s}}{100} \Rightarrow \quad V_{s}=300 \mathrm{~V}
\end{aligned}$
Hence, the correct answer is option: 'B'.